A coin is kept at distance of 10 cm from the centre of a circular turn table. If `mu=0.8` the frequency of rotation at which the coin just begins to slip is

Options

A. 62.8 rpm
B. 84.54 rpm
C. 54.6 rpm
D. 32.4 rpm

Answer

To solve the problem of finding the frequency of rotation at which the coin just begins to slip, we can follow these steps:

### Step 1: Identify the given values
- Distance from the center of the circular turntable (r) = 10 cm = 0.1 m
- Coefficient of friction (μ) = 0.8
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Write the equation for the forces
The frictional force provides the necessary centripetal force to keep the coin moving in a circle. The equation is:
\[ \mu mg = m \omega^2 r \]
Here, \( m \) is the mass of the coin, which will cancel out from both sides.

### Step 3: Simplify the equation
Cancel \( m \) from both sides:
\[ \mu g = \omega^2 r \]

### Step 4: Solve for angular velocity (ω)
Rearranging the equation gives us:
\[ \omega^2 = \frac{\mu g}{r} \]
Taking the square root:
\[ \omega = \sqrt{\frac{\mu g}{r}} \]

### Step 5: Substitute the known values
Substituting the values of μ, g, and r into the equation:
\[ \omega = \sqrt{\frac{0.8 \times 9.8}{0.1}} \]

### Step 6: Calculate ω
Calculating the value:
\[ \omega = \sqrt{\frac{7.84}{0.1}} \]
\[ \omega = \sqrt{78.4} \]
\[ \omega \approx 8.85 \text{ rad/s} \]

### Step 7: Convert angular velocity to frequency (f)
We know that:
\[ \omega = 2 \pi f \]
Rearranging gives:
\[ f = \frac{\omega}{2 \pi} \]

### Step 8: Substitute the value of ω
Substituting the value of ω:
\[ f = \frac{8.85}{2 \pi} \]
\[ f \approx \frac{8.85}{6.2832} \]
\[ f \approx 1.41 \text{ rps} \]

### Step 9: Convert frequency from rps to rpm
To convert from revolutions per second (rps) to revolutions per minute (rpm):
\[ \text{rpm} = f \times 60 \]
\[ \text{rpm} = 1.41 \times 60 \]
\[ \text{rpm} \approx 84.54 \]

### Final Answer
The frequency of rotation at which the coin just begins to slip is approximately **84.54 rpm**.

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